Regstats : Multiple regression/comparing regression slopes

Hi,

I want to compare the regression slopes of 3 data sets. Apparently you have to set some dummy variables =0/1 and regress against them to do this.

So I tried to do it using regstats but I can't get regstats to work with several regressors.

I tried the following :

X =
% the "real" regressor (1 2 3) repeated 3 times (one for each set)
     1
     2
     3
     1
     2
     3
     1
     2
     3
Z12 =
% the first dummy regressor (1 if data is from set number 2)
     0
     0
     0
     1
     1
     1
     0
     0
     0
Z20 =
% second dummy regressor (1 if data is from group 3)
     0
     0
     0
     0
     0
     0
     1
     1
     1

V=[X,Z12,Z20]
V =
% all three regressors put together in one matrix
     1     0     0
     2     0     0
     3     0     0
     1     1     0
     2     1     0
     3     1     0
     1     0     1
     2     0     1
     3     0     1

Y=2+3*X+4*Z12+5*Z20*X' 
Y =
% fake dataset to see if regstats would give me the right coefficients
    35
    38
    41
    39
    42
    45
    35
    38
    41

then i tried

 s=regstats(Y,V,'interactions','all')

and got the following warnings

Warning: Matrix is singular to working precision. 
> In regstats at 180
Warning: Matrix is singular to working precision. 
> In regstats at 195

and none of the numbers is correct i get a lot of NaNs and Inf for example

>> s.beta

ans =

     NaN
     NaN
     NaN
     NaN
     NaN
     NaN
     Inf

I don't have any clue on what is wrong with the code I tried. I have never used regstats before. Any idea on what I could do ?

Thank you very much for your help !

[Oleg Komarov]

Y=2+3*X+4*Z12+5*Z20*X' doesn't give the output you wrote but

Y=2+3*X+4*Z12+5*Z20'*X does

why are you doing that? Do you have lagged values of X in your model?

[laurie]

thank you for your answer

sorry about the stray character. the output version is the right one. i don't know what a lagged value is.

i am doing this because i have 3 sets of data y1,y2, and y3, and i want to compare the slopes of their respective regressions against x. (are the slopes equal or not?).

i think my problem is very similar to this one here

http://www.mathworks.com/matlabcentral/newsreader/view_thread/270278

but i am struggling to understand the method.

[Oleg Komarov]

The thread you found is a good one, follow it thoroughly and pay attention to details, for example .* is different from *

Answer 1

[Oleg Komarov]

Your data (in copy-pastable form):

X   = [1;2;3;1;2;3;1;2;3];
Z12 = [0;0;0;1;1;1;0;0;0];
Z20 = [0;0;0;0;0;0;1;1;1];
V   = [X,Z12,Z20];

Y   = 2 + 3*X + 4*Z12 + 5*Z20'*X;

Setting the model to 'interaction' gives this matrix:

x2fx(V,'interaction')

1  1  0  0  0  0  0
1  2  0  0  0  0  0
1  3  0  0  0  0  0
1  1  1  0  1  0  0
1  2  1  0  2  0  0
1  3  1  0  3  0  0
1  1  0  1  0  1  0
1  2  0  1  0  2  0
1  3  0  1  0  3  0

Is this what you're looking for? (Not likely). To understand what's going on with the choice of the model read the documentation of x2fx, which is used by regstats.

Answer 2

[laurie]

Thanks you very much, it worked fine !! :-)

Just checking : if regstats gives a p-value > criterium for one regressor, it means that the coefficient for this regressor is probably 0 right ? Also, should'nt the criterium be adjusted to the number of regressors ?

[Oleg Komarov]

You usually read things like, the p-values are too high, thus the coefficients are not statistically significant at any conentional level (where the conventional levels are 5% or 1%).

A p-value is the (smallest) probability to observe a value that is different from the null just by chance.

Rephrasing, a p-value is the percentage of values you would erroneously accept (when they should be considered as the null).

In regressions, the p-value of single coefficients is ususally the t-test against the null of 0.

So, say you have a coefficient of 0.32 and the question you want to ask is: "is it truly 0.32 or should I consider it as 0?"

Suppose you get a p-value of 9.81%. It says that the probability of 0.32 being a 0 is 9.81%.

Now, are you willing to accept such a risk? Usually, the threshold is 5%, thus observing a p-value lower than the threshold is good sign.

DISCLAIMER: I am abusing the words truly and accept.

Truly is with respect to the assumed distribution for your parameters and you NEVER accept the NULL but you fail to reject it.